Question: Find the zeros of the function. Enter the solutions from least to greatest. $f(x) = (x - 1)^2 - 36$ $\text{lesser }x = $
Explanation: $\begin{aligned} (x - 1)^2 - 36&= 0 \\\\ (x-1)^2&=36 \\\\ \sqrt{(x-1)^2}&=\sqrt{36} \end{aligned}$ $\begin{aligned} x-1&=\pm6 \\\\ x&=\pm6+1 \\ \phantom{(x - 1)^2 - 36}& \\ x=-5&\text{ or }x=7 \end{aligned}$ In conclusion, $\begin{aligned} \text{lesser }x &= -5 \\\\ \text{greater } x &= 7 \end{aligned}$